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Calculate enthalpy of the reaction, F(...

Calculate enthalpy of the reaction,
`F_(2)O(g)+H_(2)O(g) to O_(2)(g)+2HF(g)`
if the bond energies of O-F,O-H,H-F and O=O bonds are 44, 111, 135 and 119 kcal per mole respectively.

Text Solution

Verified by Experts

We have to calculate `DeltaH` of the reaction
`F_(2)O(g)+H_(2)O(g) to O_(2)(G)+2HF (g) `
For reactants
Bond energy of 2 moles of O-F bonds `=2xx44` kcal
Bond energy of 2 moles of O-H bonds `=2xx111` kcal
For products
Bond formation energy of 1 mole of O=O bonds `=-119` kcal
Bond formation energy of 2 moles of H-F bonds `=-2xx135` kcal
Adding, we get `DeltaH` of the required equation,
`DeltaH=2xx44+2xx111-119-2xx135=-79` kcal
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F_(2)(g)+H_(2)O(l) to

Find Delta_(r )H of the reaction : OF_(2)(g)+H_(2)O(g)rarr O_(2)(g)+2HF(g) , Average bond energies of O-F, O-H, O=O, H-F are 44, 111, 118 and 270 kcal/mol respectively.

Knowledge Check

  • The enthalpy change for the reaction- H_2(g) + (1)/(2) O_2(g) to H_2O (g) is called

    A
    Enthalpy of formation of water
    B
    Enthalpy of solution
    C
    Enthalpy of vaporisation of water
    D
    None of these

  • DeltaH for the reaction, OF_(2)+H_(2)OrarrO_(2)+2HF (B.E. of O-F, O-H, H-F and O=O are 44, 111, 135 and "119 kcal mol"^(-1) respectively)

    A
    `-222" kcal"`
    B
    `-"88 kcal"`
    C
    `-"111 kcal"`
    D
    `-"79 kcal"`

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